乘风破浪:LeetCode真题_029_Divide Two Integers 一.前言 两个整数相除,不能使用乘法除法和取余运算.那么就只能想想移位运算和加减法运算了. 二.Divide Two Integers 2.1 问题 2.2 分析与解决 通过分析,我们可以想到,如果使用加法,一次次的减下去,每减一次就加一,直到最后的减数小于除数.但是这样的时间复杂度将会是非常的大,比如100000,3这两个数,非常的耗时,那么如何加快运算呢,我们想到了移位运算. public cl…
leetcode面试准备:Divide Two Integers 1 题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 接口: public int divide(int dividend, int divisor) 2 思路 题意 不用乘.除.mod 做一个除法运算. 解 直接用除数去一个一个加,直到被除数被超过的话…
Divide two integers without using multiplication, division and mod operator. 分析:题目意思很容易理解,就是不用乘除法和模运算求来做除法,很容易想到的一个方法是一直做减法,然后计数,超时.在网上找到一种解法,利用位运算,意思是任何一个整数可以表示成以2的幂为底的一组基的线性组合,即num=a_0*2^0+a_1*2^1+a_2*2^2+...+a_n*2^n.基于以上这个公式以及左移一位相当于乘以2,我们先让除数左移直到…
一天一道LeetCode系列 (一)题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. (二)解题 这题看起来很简单,一开始想到的方法就是从0开始一次累加除数,一直到比被除数大为止,好无悬念,这样做的结果就是超时了. 用移位来实现除法效率就比较高了.具体思路可以参考二进制除法.下面举个例子来说明. 例如:10/2 即10…
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 29: Divide Two Integershttps://oj.leetcode.com/problems/divide-two-integers/ Divide two integers without using multiplication, division and mod operator.If it is overflow, return MAX_INT. =…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 两数相除,不能用*,/,% 那么只能用位移运算了,注意边界条件防止溢出,代码如下: class Solution { public: int divide(int dividend, int divisor) { ? : -; ? : -; long long d1 = a…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 题解: 思路就是被除数减去除数,减尽为止.优化的方法是尽量少的做减法.由于不能用乘法,可以利用位操作,左移一位即为该数乘上2 Solution 1 class Solution { public: int divide(int dividend, int divisor) {…
50. Pow(x, n) (中等) double myPow(double x, int n) { ; unsigned long long p; ) { p = -n; x = / x; } else { p = n; } while (p) { ) ans *= x; x *= x; p >>= ; } return ans; } 96. Unique Binary Search Trees(很快) class Solution { public: int numTrees(int n)…
题目一: 反转从位置 m 到 n 的链表.请使用一趟扫描完成反转. 说明:1 ≤ m ≤ n ≤ 链表长度. 示例: 输入: 1->2->3->4->5->NULL, m = 2, n = 4输出: 1->4->3->2->5->NULL 方法一: 1.和链表反转很相似,增加了难度,需要选择一定的区间: 2.找到区间进行反转,最后再将链表和之前的进行连接: 3.首先找到反转链表的前一位,for(int i=0;i<m-1;i++) pre=…
# Title Solution Acceptance Difficulty Frequency 1 Two Sum 44.5% Easy 2 Add Two Numbers 31.7% Medium 3 Longest Substring Without Repeating Characters 28.8% Medium 4 Median of Two Sorted Arrays 27.2% Hard 5…
题目:两整数相除 难度:Medium 题目内容: Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. 翻…
题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT 链接 https://leetcode.com/problems/divide-two-integers/ 答案 1.int的最大值MAX_INT为power(2,31)-1 = 2147483647 2.int的最小值MIN_INT为-power(2,31) = -21…
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Inp…
Divide two integers without using multiplication, division and mod operator. 题解:要求不用乘除和取模运算实现两个数的除法. 那么用加减法是很自然的选择.不过如果一次只从被除数中剪掉一个除数会TLE.所以我们借助移位运算,依次从被除数中减去1个除数,2个除数,4个除数......当减不动的时候,再依次从被除数中减去......4个除数,2个除数,1个除数. 例如50除以5的计算过程如下: dividend exp tem…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解题思路: 模拟除法运算,但是不能试用 * / % 操作. 先思考边界条件,再开始做题: 0.除数为0无意义,需要和出题人沟通此边界是否存在(leetcode中并未考察除数为0的情况,默认除数不为0): 1.传入被除数/除数两个int做除法,什么情况返回值overflow:被…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 这道题让我们求两数相除,而且规定我们不能用乘法,除法和取余操作,那么我们还可以用另一神器位操作Bit Operation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更…
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: divide…
Divide Two Integers Divide two integers without using multiplication, division and mod operator. SOLUTION 11. 基本思想是不断地减掉除数,直到为0为止.但是这样会太慢. 2. 我们可以使用2分法来加速这个过程.不断对除数*2,直到它比被除数还大为止.加倍的同时,也记录下cnt,将被除数减掉加倍后的值,并且结果+cnt. 因为是2倍地加大,所以速度会很快,指数级的速度. 3. 另外要注意的是…
1. Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the…
Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator. Return the quotient after dividing dividend by divisor. The integer division should truncate toward zero. Example 1: Input: dividend…
Divide two integers without using multiplication, division and mod operator. 不用乘.除.求余操作,返回两整数相除的结果,结果也是整数. 假设除数是2,相除的商就是被除数二进制表示向右移动一位. 假设被除数是a,除数是b,因为不知道a除以b的商,所以只能从b,2b,4b,8b.......这种序列一个个尝试 从a扣除那些尝试的值. 如果a大于序列的数,那么a扣除该值,并且最终结果是商加上对应的二进制位为1的数,然后尝试序…
题目:Divide Two Integers Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 看讨论区大神的思路: In this problem, we are asked to divide two integers. However, we are not allowed to use division, multi…
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 解题思路: 既然不呢个用乘除和取模运算,只好采用移位运算,可以通过设置一个length代表divisor向左的做大移位数,直到大于dividend,然后对length做一个循环递减,dividend如果大于divisor即进行减法运算,同时result加上对应的值,注意边界条…
